ex00: ft_strlen

Repeat of C01 ex06

We create an int i and start at 0, then while we don't have a \0 in str[i] keep incrementing i to move forward.

At the end, i is the length of our string since we finished walking it.

ex01: ft_putstr

While the current char that *str points at isn't \0, use write to write said character and move *str forward by incrementing it.

ex02: ft_putnbr

Repeat of C00 ex07

I solved this one through recursion.

Basically when the function is first run, it checks if the number entered is -2147483648 because this is the overflow point for int, so it would cause unintended behavior if we try to calculate it. For this special case number we manually do the write.

If it is not, then we go on to check if the number is negative, and if it is we need to put a dash so it shows up as negative, then we set nb = -nb which basically makes it positive so we can work with it, otherwise the algorithm will fail.

Then if the number only has 1 digit (more or equal to 0 and less than 10) we can easily reuse ft_putchar again to output that. Otherwise, we will call ft_putnbr again, this time dividing nb by 10 so we remove the rightmost digit and do the same procedure on top of it again, and this keeps repeating until we only have one digit we can output using ft_putchar. Then after all these recursive operations, it does the same but this time using the % operator, which does the opposite of division by 10: it removes the leftmost digit. And this way, both halves are eventually reduced to single characters that can be outputted using ft_putchar, leading to our number!

ex03: ft_atoi

We start by defining 2 int:

• res for storing the result so far
• neg stores a number thats either 1 or -1, that gets multiplied by res as the result in the end to support negative numbers.

We skip spaces and other empty ASCII chars by incrementing the pointer while it still is these

Then we go through the + and - signs. If +, we simply ignore it, if - we invert neg, so 1 becomes -1 or -1 becomes 1.

And then while we have number char remaining, we add them to res as follows:

while (*str >= '0' && *str <= '9')
{
	res = res * 10 + (*str - '0');
	str++;
}

Finally, we return res times neg as the result.

ex04: ft_putnbr_base

Similar approach to ft_putnbr except we also have to convert decimal to a new base. We first check the base with 42's rules, so it shouldn't contain +, - or space and no duplicate characters, and at least 2 characters long. If all of that passes we also check if the number is negative and print - if it is, then we make it positive again to do the math properly. Now instead of dividing by 10 we divide by our base length and we choose the digits from the base instead of decimal digits.

ex05: ft_atoi_base

We do basically the same checks to the base we did in ft_putnbr_base in the beginning.

Using ft_find_dval we find the value of a digit inside the str (and that's basically the index of the found digit in the base) and then add it if found. Once it is no longer found, that's our end value.

Browse source files 6 files in C04. They open in Notepad.

This page is C04's notes, rendered here from the README on GitHub. Last read on 18 October 2026 at 16:57 UTC.